If $f(x)=int_{3}^x sqrt{1+t^3} dt$ Find $(f^{-1})'(0)$












2












$begingroup$


Consider the integral $$f(x)=int_{3}^x sqrt{1+t^3} dt$$



Using Theorem 7 from my textbook which is $$frac{1}{f'(f^{-1}(a))}$$



$f'(t)= frac{2t^3}{2sqrt{1+t^3}}$



$f^{-1}(t)=sqrt[3]{t^2-1}$



I get confused here because does the integral affect how I should proceed or should I have incorporated the integral from the start?






calculus inverse-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your derivative step is very wrong; a big red flag is the fact that you have a function of $x$ equal to a function of $t$.
    $endgroup$
    – T. Bongers
    Dec 10 '18 at 23:49










  • $begingroup$
    It was a force of habit, i'm used to $f(x)$
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 23:51










  • $begingroup$
    It's still quite wrong. You've differentiated $sqrt{1 + t^3}$, not $f$. You've also just changed where the notation is incorrect to the previous line!
    $endgroup$
    – T. Bongers
    Dec 10 '18 at 23:52










  • $begingroup$
    Yeah so people like you don't comment on my tiny mistakes. I'm sleep deprived and over stressed, give me a break.
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 23:55








  • 1




    $begingroup$
    The very first expression after “Consider the integral” is wrong: $f(t)$ should be $f(x)$. Also (here and in real life) I highly recommend that you don’t be rude to someone who’s trying to help you.
    $endgroup$
    – Shalop
    Dec 11 '18 at 0:13


















2












$begingroup$


Consider the integral $$f(x)=int_{3}^x sqrt{1+t^3} dt$$



Using Theorem 7 from my textbook which is $$frac{1}{f'(f^{-1}(a))}$$



$f'(t)= frac{2t^3}{2sqrt{1+t^3}}$



$f^{-1}(t)=sqrt[3]{t^2-1}$



I get confused here because does the integral affect how I should proceed or should I have incorporated the integral from the start?






calculus inverse-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your derivative step is very wrong; a big red flag is the fact that you have a function of $x$ equal to a function of $t$.
    $endgroup$
    – T. Bongers
    Dec 10 '18 at 23:49










  • $begingroup$
    It was a force of habit, i'm used to $f(x)$
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 23:51










  • $begingroup$
    It's still quite wrong. You've differentiated $sqrt{1 + t^3}$, not $f$. You've also just changed where the notation is incorrect to the previous line!
    $endgroup$
    – T. Bongers
    Dec 10 '18 at 23:52










  • $begingroup$
    Yeah so people like you don't comment on my tiny mistakes. I'm sleep deprived and over stressed, give me a break.
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 23:55








  • 1




    $begingroup$
    The very first expression after “Consider the integral” is wrong: $f(t)$ should be $f(x)$. Also (here and in real life) I highly recommend that you don’t be rude to someone who’s trying to help you.
    $endgroup$
    – Shalop
    Dec 11 '18 at 0:13
















2












2








2





$begingroup$


Consider the integral $$f(x)=int_{3}^x sqrt{1+t^3} dt$$



Using Theorem 7 from my textbook which is $$frac{1}{f'(f^{-1}(a))}$$



$f'(t)= frac{2t^3}{2sqrt{1+t^3}}$



$f^{-1}(t)=sqrt[3]{t^2-1}$



I get confused here because does the integral affect how I should proceed or should I have incorporated the integral from the start?






calculus inverse-function







share|cite|improve this question











$endgroup$




Consider the integral $$f(x)=int_{3}^x sqrt{1+t^3} dt$$



Using Theorem 7 from my textbook which is $$frac{1}{f'(f^{-1}(a))}$$



$f'(t)= frac{2t^3}{2sqrt{1+t^3}}$



$f^{-1}(t)=sqrt[3]{t^2-1}$



I get confused here because does the integral affect how I should proceed or should I have incorporated the integral from the start?






calculus inverse-function




calculus inverse-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 0:15







Eric Brown

















asked Dec 10 '18 at 23:39









Eric BrownEric Brown

737




737












  • $begingroup$
    Your derivative step is very wrong; a big red flag is the fact that you have a function of $x$ equal to a function of $t$.
    $endgroup$
    – T. Bongers
    Dec 10 '18 at 23:49










  • $begingroup$
    It was a force of habit, i'm used to $f(x)$
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 23:51










  • $begingroup$
    It's still quite wrong. You've differentiated $sqrt{1 + t^3}$, not $f$. You've also just changed where the notation is incorrect to the previous line!
    $endgroup$
    – T. Bongers
    Dec 10 '18 at 23:52










  • $begingroup$
    Yeah so people like you don't comment on my tiny mistakes. I'm sleep deprived and over stressed, give me a break.
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 23:55








  • 1




    $begingroup$
    The very first expression after “Consider the integral” is wrong: $f(t)$ should be $f(x)$. Also (here and in real life) I highly recommend that you don’t be rude to someone who’s trying to help you.
    $endgroup$
    – Shalop
    Dec 11 '18 at 0:13




















  • $begingroup$
    Your derivative step is very wrong; a big red flag is the fact that you have a function of $x$ equal to a function of $t$.
    $endgroup$
    – T. Bongers
    Dec 10 '18 at 23:49










  • $begingroup$
    It was a force of habit, i'm used to $f(x)$
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 23:51










  • $begingroup$
    It's still quite wrong. You've differentiated $sqrt{1 + t^3}$, not $f$. You've also just changed where the notation is incorrect to the previous line!
    $endgroup$
    – T. Bongers
    Dec 10 '18 at 23:52










  • $begingroup$
    Yeah so people like you don't comment on my tiny mistakes. I'm sleep deprived and over stressed, give me a break.
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 23:55








  • 1




    $begingroup$
    The very first expression after “Consider the integral” is wrong: $f(t)$ should be $f(x)$. Also (here and in real life) I highly recommend that you don’t be rude to someone who’s trying to help you.
    $endgroup$
    – Shalop
    Dec 11 '18 at 0:13


















$begingroup$
Your derivative step is very wrong; a big red flag is the fact that you have a function of $x$ equal to a function of $t$.
$endgroup$
– T. Bongers
Dec 10 '18 at 23:49




$begingroup$
Your derivative step is very wrong; a big red flag is the fact that you have a function of $x$ equal to a function of $t$.
$endgroup$
– T. Bongers
Dec 10 '18 at 23:49












$begingroup$
It was a force of habit, i'm used to $f(x)$
$endgroup$
– Eric Brown
Dec 10 '18 at 23:51




$begingroup$
It was a force of habit, i'm used to $f(x)$
$endgroup$
– Eric Brown
Dec 10 '18 at 23:51












$begingroup$
It's still quite wrong. You've differentiated $sqrt{1 + t^3}$, not $f$. You've also just changed where the notation is incorrect to the previous line!
$endgroup$
– T. Bongers
Dec 10 '18 at 23:52




$begingroup$
It's still quite wrong. You've differentiated $sqrt{1 + t^3}$, not $f$. You've also just changed where the notation is incorrect to the previous line!
$endgroup$
– T. Bongers
Dec 10 '18 at 23:52












$begingroup$
Yeah so people like you don't comment on my tiny mistakes. I'm sleep deprived and over stressed, give me a break.
$endgroup$
– Eric Brown
Dec 10 '18 at 23:55






$begingroup$
Yeah so people like you don't comment on my tiny mistakes. I'm sleep deprived and over stressed, give me a break.
$endgroup$
– Eric Brown
Dec 10 '18 at 23:55






1




1




$begingroup$
The very first expression after “Consider the integral” is wrong: $f(t)$ should be $f(x)$. Also (here and in real life) I highly recommend that you don’t be rude to someone who’s trying to help you.
$endgroup$
– Shalop
Dec 11 '18 at 0:13






$begingroup$
The very first expression after “Consider the integral” is wrong: $f(t)$ should be $f(x)$. Also (here and in real life) I highly recommend that you don’t be rude to someone who’s trying to help you.
$endgroup$
– Shalop
Dec 11 '18 at 0:13












1 Answer
1






active

oldest

votes


















4












$begingroup$

HINTS:



From the Fundamental Theorem of Calculus, we have



$$f'(x)=sqrt{1+x^3}$$



Moreover, note that



$$f(3)=0implies f^{-1}(0)=3$$



Finally, the theorem from your text book reveals that



$$frac{df^{-1}(x)}{dx}=frac1{sqrt{1+(f^{-1}(x))^3}}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is my inverse function correct though?
    $endgroup$
    – Eric Brown
    Dec 11 '18 at 0:04










  • $begingroup$
    Hi Eric. No, the inverse of $int_3^xsqrt{1+t^3},dt$ is not $sqrt[3]{x^2-1}$ .
    $endgroup$
    – Mark Viola
    Dec 11 '18 at 0:11











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

HINTS:



From the Fundamental Theorem of Calculus, we have



$$f'(x)=sqrt{1+x^3}$$



Moreover, note that



$$f(3)=0implies f^{-1}(0)=3$$



Finally, the theorem from your text book reveals that



$$frac{df^{-1}(x)}{dx}=frac1{sqrt{1+(f^{-1}(x))^3}}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is my inverse function correct though?
    $endgroup$
    – Eric Brown
    Dec 11 '18 at 0:04










  • $begingroup$
    Hi Eric. No, the inverse of $int_3^xsqrt{1+t^3},dt$ is not $sqrt[3]{x^2-1}$ .
    $endgroup$
    – Mark Viola
    Dec 11 '18 at 0:11
















4












$begingroup$

HINTS:



From the Fundamental Theorem of Calculus, we have



$$f'(x)=sqrt{1+x^3}$$



Moreover, note that



$$f(3)=0implies f^{-1}(0)=3$$



Finally, the theorem from your text book reveals that



$$frac{df^{-1}(x)}{dx}=frac1{sqrt{1+(f^{-1}(x))^3}}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is my inverse function correct though?
    $endgroup$
    – Eric Brown
    Dec 11 '18 at 0:04










  • $begingroup$
    Hi Eric. No, the inverse of $int_3^xsqrt{1+t^3},dt$ is not $sqrt[3]{x^2-1}$ .
    $endgroup$
    – Mark Viola
    Dec 11 '18 at 0:11














4












4








4





$begingroup$

HINTS:



From the Fundamental Theorem of Calculus, we have



$$f'(x)=sqrt{1+x^3}$$



Moreover, note that



$$f(3)=0implies f^{-1}(0)=3$$



Finally, the theorem from your text book reveals that



$$frac{df^{-1}(x)}{dx}=frac1{sqrt{1+(f^{-1}(x))^3}}$$






share|cite|improve this answer









$endgroup$



HINTS:



From the Fundamental Theorem of Calculus, we have



$$f'(x)=sqrt{1+x^3}$$



Moreover, note that



$$f(3)=0implies f^{-1}(0)=3$$



Finally, the theorem from your text book reveals that



$$frac{df^{-1}(x)}{dx}=frac1{sqrt{1+(f^{-1}(x))^3}}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 23:53









Mark ViolaMark Viola

131k1275171




131k1275171












  • $begingroup$
    Is my inverse function correct though?
    $endgroup$
    – Eric Brown
    Dec 11 '18 at 0:04










  • $begingroup$
    Hi Eric. No, the inverse of $int_3^xsqrt{1+t^3},dt$ is not $sqrt[3]{x^2-1}$ .
    $endgroup$
    – Mark Viola
    Dec 11 '18 at 0:11


















  • $begingroup$
    Is my inverse function correct though?
    $endgroup$
    – Eric Brown
    Dec 11 '18 at 0:04










  • $begingroup$
    Hi Eric. No, the inverse of $int_3^xsqrt{1+t^3},dt$ is not $sqrt[3]{x^2-1}$ .
    $endgroup$
    – Mark Viola
    Dec 11 '18 at 0:11
















$begingroup$
Is my inverse function correct though?
$endgroup$
– Eric Brown
Dec 11 '18 at 0:04




$begingroup$
Is my inverse function correct though?
$endgroup$
– Eric Brown
Dec 11 '18 at 0:04












$begingroup$
Hi Eric. No, the inverse of $int_3^xsqrt{1+t^3},dt$ is not $sqrt[3]{x^2-1}$ .
$endgroup$
– Mark Viola
Dec 11 '18 at 0:11




$begingroup$
Hi Eric. No, the inverse of $int_3^xsqrt{1+t^3},dt$ is not $sqrt[3]{x^2-1}$ .
$endgroup$
– Mark Viola
Dec 11 '18 at 0:11


















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